Question

question 4
solve for x.
$x^2 + 6x + 6 = 0$
$\bigcirc -3 \pm \sqrt{15}$
$\bigcirc 3 \pm \sqrt{3}$
$\bigcirc -3 \pm \sqrt{3}$
$\bigcirc 3 \pm \sqrt{15}$

Explanation:

Step1: Identify quadratic formula

The quadratic formula for \(ax^2 + bx + c = 0\) is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
For \(x^2 + 6x + 6 = 0\), \(a = 1\), \(b = 6\), \(c = 6\).

Step2: Calculate discriminant (\(b^2 - 4ac\))

Substitute values: \(6^2 - 4(1)(6) = 36 - 24 = 12\)? Wait, no—wait, \(b^2 - 4ac = 36 - 24 = 12\)? Wait, no, wait: \(a=1\), \(b=6\), \(c=6\). So \(b^2 - 4ac = 36 - 24 = 12\)? Wait, but the options have \(\sqrt{3}\) or \(\sqrt{15}\). Wait, no, maybe I miscalculated. Wait, \(x^2 + 6x + 6 = 0\). Let's complete the square.

Step3: Complete the square

\(x^2 + 6x = -6\)
Add \((\frac{6}{2})^2 = 9\) to both sides: \(x^2 + 6x + 9 = -6 + 9\)
\((x + 3)^2 = 3\)
Take square roots: \(x + 3 = \pm \sqrt{3}\)
Subtract 3: \(x = -3 \pm \sqrt{3}\)

Wait, but earlier discriminant: \(b^2 - 4ac = 36 - 24 = 12\)? No, wait, \(c = 6\), so \(4ac = 24\), \(36 - 24 = 12\). But completing the square gave 3. Wait, no—wait, \(x^2 + 6x + 6 = 0\). Completing the square: \(x^2 + 6x = -6\). Add 9: \(x^2 + 6x + 9 = 3\), so \((x + 3)^2 = 3\). So \(x = -3 \pm \sqrt{3}\). Ah, right, because \(9 - 6 = 3\), not 12. Wait, my mistake earlier: \(c = 6\), so when moving to the other side, it's \(-6\), then adding 9 gives 3. So discriminant: \(b^2 - 4ac = 36 - 24 = 12\)? Wait, no, that's a contradiction. Wait, no: \(a=1\), \(b=6\), \(c=6\). So \(b^2 - 4ac = 36 - 24 = 12\). But completing the square gave \((x + 3)^2 = 3\), which would imply discriminant is 12? Wait, no, \((x + 3)^2 = 3\) expands to \(x^2 + 6x + 9 = 3\), so \(x^2 + 6x + 6 = 0\). So discriminant is \(36 - 24 = 12\), but \(\sqrt{12} = 2\sqrt{3}\), so when using quadratic formula: \(x = \frac{-6 \pm \sqrt{12}}{2} = \frac{-6 \pm 2\sqrt{3}}{2} = -3 \pm \sqrt{3}\). Ah, there we go. So \(\sqrt{12} = 2\sqrt{3}\), so dividing by 2 gives \(\sqrt{3}\). So the solution is \(x = -3 \pm \sqrt{3}\), which is option C (the third option: \(-3 \pm \sqrt{3}\)).

Answer:

\(-3 \pm \sqrt{3}\) (corresponding to the option with \(-3 \pm \sqrt{3}\))